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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
3장, 문제 31

Find and simplify the derivative of the following functions.
h(x) = (x − 1)(x3+ x2 + x+1)

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Step 1: Identify the function h(x) = (x - 1)(x^3 + x^2 + x + 1). This is a product of two functions, so we will use the product rule to find the derivative.
Step 2: Recall the product rule for derivatives: if u(x) and v(x) are functions, then the derivative of their product is given by (uv)' = u'v + uv'.
Step 3: Let u(x) = x - 1 and v(x) = x^3 + x^2 + x + 1. First, find the derivative of u(x), which is u'(x) = 1.
Step 4: Next, find the derivative of v(x). Differentiate each term separately: v'(x) = 3x^2 + 2x + 1.
Step 5: Apply the product rule: h'(x) = u'(x)v(x) + u(x)v'(x). Substitute the derivatives and original functions: h'(x) = (1)(x^3 + x^2 + x + 1) + (x - 1)(3x^2 + 2x + 1). Simplify the expression to find the derivative.

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이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
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4m

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Derivative

The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. The derivative is a fundamental concept in calculus, representing the slope of the tangent line to the curve of the function at any given point.
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Product Rule

The product rule is a formula used to find the derivative of the product of two functions. It states that if you have two functions, u(x) and v(x), the derivative of their product is given by u'v + uv'. This rule is essential when differentiating functions that are expressed as products, such as the function h(x) in the question.
추천 영상:
05:18
The Product Rule

Simplification of Derivatives

After finding the derivative of a function, simplification is often necessary to express the result in a more manageable form. This may involve combining like terms, factoring, or reducing fractions. Simplifying the derivative helps in understanding the behavior of the function, such as identifying critical points and analyzing concavity.
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